using trigonometry to prove that a function is less than or equal to sqrt(2)

hearts123

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Feb 22, 2019
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Hey everyone! Here's a question which states that I have to solve it with trigonometry. I had the idea of substituting x as sine or cosine, because of the range but I don't know which would be more suitable. I'm supposed to substitute trigonometric identities (like sin α or something and have a range for α) to do the question. Please help asap!

f(x)= ax+b
2a^2+ 6b^2 =3
x belongs to: [-1, 1]

Prove: |f(x)| is less than or equal to sqrt(2)
 

HallsofIvy

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Jan 27, 2012
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First, divide both sides of \(\displaystyle 2a^2+ 6b^2= 3\) by 3 to get \(\displaystyle \frac{2}{3}a^2+ 2b^2= \left(\sqrt{\frac{2}{3}}a\right)^2+ \left(\sqrt{2}b\right)^2= 1\). Now let \(\displaystyle sin(\theta)= \sqrt{\frac{2}{3}}a\), so that \(\displaystyle a= \sqrt{\frac{3}{2}}sin(\theta)\) and \(\displaystyle cos(\theta)= \sqrt{2}b\), so that \(\displaystyle b= \frac{\sqrt{2}}{2}cos(\theta)\) (or vice-versa- it doesn't matter). What can you say about \(\displaystyle f(x)= ax+ b= \sqrt{\frac{3}{2}}sin(\theta)x+ \frac{\sqrt{2}}{2}cos(\theta)\) with x between -1 and 1 and \(\displaystyle \theta\) between 0 and \(\displaystyle 2\pi\)?
 

hearts123

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Feb 22, 2019
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First, divide both sides of \(\displaystyle 2a^2+ 6b^2= 3\) by 3 to get \(\displaystyle \frac{2}{3}a^2+ 2b^2= \left(\sqrt{\frac{2}{3}}a\right)^2+ \left(\sqrt{2}b\right)^2= 1\). Now let \(\displaystyle sin(\theta)= \sqrt{\frac{2}{3}}a\), so that \(\displaystyle a= \sqrt{\frac{3}{2}}sin(\theta)\) and \(\displaystyle cos(\theta)= \sqrt{2}b\), so that \(\displaystyle b= \frac{\sqrt{2}}{2}cos(\theta)\) (or vice-versa- it doesn't matter). What can you say about \(\displaystyle f(x)= ax+ b= \sqrt{\frac{3}{2}}sin(\theta)x+ \frac{\sqrt{2}}{2}cos(\theta)\) with x between -1 and 1 and \(\displaystyle \theta\) between 0 and \(\displaystyle 2\pi\)?
Am I right in saying that the maximum possible value is sqrt(2) / 2 and minimum is - sqrt(2) / 2 ?
 

HallsofIvy

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Well, how did you get that answer?
 

Jomo

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Dec 30, 2014
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Think very carefully about the statement or vice-versa- it doesn't matter that professor HallsofIvy made in a previous post. It is a very powerful statement. It should allow you to know x and \(\displaystyle \theta\)
 
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